What is the time complexity of Count the Number of Ideal Arrays?

The optimal solution has a time complexity of O(n * maxValue * log(maxValue)).

What is the space complexity of Count the Number of Ideal Arrays?

The space complexity is O(maxValue).

What data structures are used to solve Count the Number of Ideal Arrays?

This problem uses Math, Dynamic Programming, Combinatorics, Number Theory. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count the Number of Ideal Arrays?

Insight 1: The number of distinct values in an ideal array is limited by the maximum value. Since each element must be a multiple of the previous one, the maximum number of distinct elements possible is log2(maxValue) + 1. Insight 2: The problem can be solved using dynamic programming (DP) to count the number of ways to form increasing sequences based on divisibility, and combinatorics to account for the repetition of elements within the array.

Which companies ask LeetCode 2338. Count the Number of Ideal Arrays in interviews?

This problem is frequently asked at Infosys.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Count the Number of Ideal Arrays - LeetCode 2338 Solution

Q: How to solve Count the Number of Ideal Arrays?

The solution uses dynamic programming (DP) to count the number of increasing sequences formed by the divisibility rule, combined with combinatorial calculations. The DP table `dp[i]` represents the number of ideal arrays ending with the value `i`. We iterate through the possible lengths of the ideal array (from 2 to min(n, max_len) + 1) where `max_len = log2(maxValue) + 1`. For each length, we compute the number of ways to form an ideal array of that length using divisibility. Then, we use combinations to consider the repetition of values within the entire array of length `n`.

Count the Number of Ideal Arrays - Complete Solution Guide

Count the Number of Ideal Arrays is LeetCode problem 2338, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two integers n and maxValue , which are used to describe an ideal array. A 0-indexed integer array arr of length n is considered ideal if the following conditions hold: Every arr[i] is a value from 1 to maxValue , for 0 <= i < n . Every arr[i] is divisible by arr[i - 1] , for 0 < i < n . Return the number of distinct ideal arrays of length n . Since the answer may be very large, return it modulo 10 9 + 7 . Example 1: Input: n = 2, maxValue = 5 Output: 10 Explanation: The following

Detailed Explanation

The problem asks us to count the number of "ideal arrays" of length `n` where each element is between 1 and `maxValue`. An ideal array has the property that each element is divisible by the element before it. The result should be returned modulo 10^9 + 7.

Solution Approach

The solution uses dynamic programming (DP) to count the number of increasing sequences formed by the divisibility rule, combined with combinatorial calculations. The DP table `dp[i]` represents the number of ideal arrays ending with the value `i`. We iterate through the possible lengths of the ideal array (from 2 to min(n, max_len) + 1) where `max_len = log2(maxValue) + 1`. For each length, we compute the number of ways to form an ideal array of that length using divisibility. Then, we use combinations to consider the repetition of values within the entire array of length `n`.

Step-by-Step Algorithm

Step 1: Initialize `dp` array. `dp[i] = 1` for all `i` from 1 to `maxValue`, indicating that there is one ideal array of length 1 for each value.
Step 2: Calculate `max_len = log2(maxValue) + 1`, the maximum possible length of distinct values in an ideal array.
Step 3: Iterate through the possible lengths of the ideal array, from `l = 2` to `min(n, max_len) + 1`.
Step 4: For each length `l`, calculate the combination `C(n-1, k)`, where `k = l - 1`. This represents the number of ways to choose the positions for `l - 1` distinct values within the array of length `n`.
Step 5: Construct the next `dp` array, `next_dp`. Iterate through each element `u` from 1 to `maxValue`. For each `u`, iterate through all its multiples `v` (u*2, u*3, ...) up to `maxValue`. Update `next_dp[v] += dp[u]`. This step calculates how many arrays can end with v given that the previous value was u, and u divides v.
Step 6: Update `dp = next_dp` for next iteration.
Step 7: Calculate the sum `lk` of all elements in the current `dp` array.
Step 8: Calculate the term `comb * lk % MOD` and add it to `total_ans`.
Step 9: Return `total_ans`.

Key Insights

Insight 1: The number of distinct values in an ideal array is limited by the maximum value. Since each element must be a multiple of the previous one, the maximum number of distinct elements possible is log2(maxValue) + 1.
Insight 2: The problem can be solved using dynamic programming (DP) to count the number of ways to form increasing sequences based on divisibility, and combinatorics to account for the repetition of elements within the array.
Insight 3: Pre-computing the combinations can improve efficiency. However, since `n` and `maxValue` are relatively small, calculating combinations on the fly with modular inverse is sufficient.

Complexity Analysis

Time Complexity: O(n * maxValue * log(maxValue))

Space Complexity: O(maxValue)

Topics

This problem involves: Math, Dynamic Programming, Combinatorics, Number Theory.

Companies

Asked at: Infosys.

Count the Number of Ideal Arrays | LeetCode 2338

Hard

Math Dynamic Programming Combinatorics Number Theory

Time: O(n * maxValue * log(maxValue))Space: O(maxValue)

Solution Code

Available in:

import java.lang.Math;

class Solution {
    public int idealArrays(int n, int maxValue) {
        int MOD = 1000000007;

        int max_len = (int)(Math.log(maxValue) / Math.log(2)) + 1;
        
        int[] dp = new int[maxValue + 1];
        for(int i = 1; i <= maxValue; i++){
            dp[i] = 1;
        }
        dp[0] = 0;
        
        int total_ans = maxValue;
        
        long comb = 1;

        for (int l = 2; l <= Math.min(n, max_len) + 1; l++) {
            int k = l - 1;
            if (k > n - 1) {
                break;
            }
            
            comb = comb * (n - k) % MOD;
            comb = comb * power(k, MOD - 2, MOD) % MOD;

            int[] next_dp = new int[maxValue + 1];
            for (int u = 1; u <= maxValue; u++) {
                if (dp[u] == 0) {
                    continue;
                }
                for (int v = u * 2; v <= maxValue; v += u) {
                    next_dp[v] = (next_dp[v] + dp[u]) % MOD;
                }
            }
            
            dp = next_dp;
            
            long lk = 0;
            for(int val : dp){
                lk = (lk + val) % MOD;
            }

            if (lk == 0) {
                break;
            }
            
            long term = (comb * (lk % MOD)) % MOD;
            total_ans = (total_ans + (int)term) % MOD;
            
        }
        
        return total_ans;
    }

    private long power(long base, long exp, long mod) {
        long res = 1;
        base %= mod;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % mod;
            base = (base * base) % mod;
            exp >>= 1;
        }
        return res;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2338: Count the Number of Ideal Arrays - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Combinatorics Number Theory

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2338

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize `dp` array. `dp[i] = 1` for all `i` from 1 to `maxValue`, indicating that there is one ideal array of length 1 for each value.
2
Step 2: Calculate `max_len = log2(maxValue) + 1`, the maximum possible length of distinct values in an ideal array.
3
Step 3: Iterate through the possible lengths of the ideal array, from `l = 2` to `min(n, max_len) + 1`.
4
Step 4: For each length `l`, calculate the combination `C(n-1, k)`, where `k = l - 1`. This represents the number of ways to choose the positions for `l - 1` distinct values within the array of length `n`.
5
Step 5: Construct the next `dp` array, `next_dp`. Iterate through each element `u` from 1 to `maxValue`. For each `u`, iterate through all its multiples `v` (u*2, u*3, ...) up to `maxValue`. Update `next_dp[v] += dp[u]`. This step calculates how many arrays can end with v given that the previous value was u, and u divides v.
6
Step 6: Update `dp = next_dp` for next iteration.
7
Step 7: Calculate the sum `lk` of all elements in the current `dp` array.
8
Step 8: Calculate the term `comb * lk % MOD` and add it to `total_ans`.
9
Step 9: Return `total_ans`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n * maxValue * log(maxValue))

💾

Space Complexity

O(maxValue)

📝

Why?

The nested loops dominate the runtime. The outer loop iterates `n` or `log_2(maxValue)` times, whichever is smaller. The inner loops iterate up to `maxValue`, and for each number, we iterate through its multiples, which takes roughly `log(maxValue)` time on average.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach can be used to explore the possible arrays. Memoization can store the number of arrays found for intermediate states to avoid redundant calculations.

Time: O(n * maxValue)Space: O(n * maxValue)

Trade-offs: Recursion with memoization might be conceptually simpler, but due to function call overhead it may be slightly slower than iterative DP, especially for large values of `n` and `maxValue`. Space complexity is also higher due to the recursion stack, potentially leading to stack overflow issues.

✨

Key Insights

1Insight 1: The number of distinct values in an ideal array is limited by the maximum value. Since each element must be a multiple of the previous one, the maximum number of distinct elements possible is log2(maxValue) + 1.
2Insight 2: The problem can be solved using dynamic programming (DP) to count the number of ways to form increasing sequences based on divisibility, and combinatorics to account for the repetition of elements within the array.
3Insight 3: Pre-computing the combinations can improve efficiency. However, since `n` and `maxValue` are relatively small, calculating combinations on the fly with modular inverse is sufficient.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `n = 2`, `maxValue = 1`. The correct answer is 1. The code correctly handles this since the max_len is at least 1 and the loop will calculate the correct amount.
Edge case 2: `maxValue = 1`. There is only one possible value in the array, so the answer is 1. The initialization of `total_ans = maxValue` handles this case.
Edge case 3: `n` is large compared to `maxValue`. The outer loop iterates only up to `log2(maxValue) + 1` because that's the maximum number of distinct factors an ideal array can have.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step. This can lead to integer overflow and incorrect results.

✗Common mistake 2: Incorrectly calculating combinations. Understanding the formula and implementing it correctly with modular inverse is crucial.

✗Common mistake 3: Not considering the maximum number of distinct values, or computing it incorrectly. For large `n` compared to `maxValue`, the outer loop should only iterate up to the maximum possible number of distinct values, which is `log2(maxValue) + 1`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Counting valid sequences based on divisibility in financial modeling or data analysis.

🏦Real-world scenario 2: Optimizing database queries where certain values must be factors of others, such as product categories and subcategories.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count the Number of Ideal Arrays"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?